Lecture Applied econometrics course - Chapter 3: Statistic inference and hypothesis testing

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Lecture Applied econometrics course - Chapter 3: Statistic inference and hypothesis testing

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Lecture Applied econometrics course - Chapter 3: Statistic inference and hypothesis testing has content: The distribution of the parameters, hypotheses testing, testing multiple linear restrictions,... and other contents.

APPLIED ECONOMETRICS COURSE Chapter III Statistic Inference and Hypothesis Testing NGUYEN BA TRUNG ­ 2016 I THE DISTRIBUTION OF THE PARAMETERS Assumpt 6. Ui ~ N(0,  2) Theorem 4.1: Normal Distribution of the  parametors  ˆ −β β βˆ0 ~ N ( β , σ ) � Z = ~ N (0,1) σ βˆ βˆ0 ˆ −β β βˆ1 ~ N ( β1 , σ ) � Z = 1 ~ N (0,1) σ βˆ βˆ1 NGUYEN BA TRUNG ­ 2016 I THE DISTRIBUTION OF THE PARAMETERS • Under the assumptions from 1-6, we have the following theorem: Theorem 4.2: Student Distribution βˆ j − β j t= ~ t(n − k − 1) se( βˆ ) j NGUYEN BA TRUNG ­ 2016 II HYPOTHESES TESTING 2.1. Testing against One­Sided Alternatives:  Right Hand Side H : β1 = (1) Manifest the hypothesis: H1 : β1 > βˆ1 − β1 (2). Compute t­statistics: t = se( βˆ ) (3).  Search  to  find  t  (n­k­1),  for  example,    α=5%,  n­k­1=28,    t0.05  (28)=1.701 ( excel: TINV(0.1,28))  (4). Decision Rule:  Nếu t >  t  (n­k­1) thì có thể bác bo gia thiê ̉ ̉ ́t H0 Nếu  t  NGUYEN BA TRUNG ­ 2016 II HYPOTHESES TESTING βˆ1 − β1 0.509 (2). Compute s­statistics: t = se( βˆ ) = 0.035 = 14.54 (3). Search to find: t0.05 (8) = 1.86 (4). Decision Rule: t­statistics = 14.54  >  t0.05 (8)=1.86, we therefore  can reject the null hypothesis H0. In other word, we can accept the  alternative hypothesis H1:  1 > 0 NGUYEN BA TRUNG ­ 2016 II. HYPOTHESES TESTING  2.2 Testing against One-Sided Alternatives: Left Hand (1) Side Suppose the hypothesis: H : β1 = H1 : β1 < (2).Compute t­statistics: βˆ1 − β1 t= se( βˆ1 ) (3).Search to find t  (n­k­1) (4). Decision Rule :  Nếu t   t /2(n­k­1), we can reject the null hypothesis H0 if   t     t /2(n­k­1), we can accept the null hypothesis H0 NGUYEN BA TRUNG ­ 2016 Example: Whether income affects to expenditure? Solution: (1) Manifest the hypothesis: H : β1 = H1 : β1 (2) Compute s-statistics: βˆ − β 0.5091 − t= = = 14.243 0.035742 se( βˆ1 ) (3) With = 5%, we have: t0.025(n-k-1) = t0.025(8) = 2.306 (4) Because t =14.243 > t0.025(8) = 2.36, we can reject the null hypothesis H0: β1= This means that income significantly impacts to expenditure NGUYEN BA TRUNG ­ 2016 We can utilize confident interval in testing hypothesis, for instance • H : β1 = 0.3 H1 : β1 v 0.3  The confident interval for  1:   (0.4268  F0.05 (1,8)= 5.32, we can reject H0 v EVIEWs provide the p­value of F­test(next slide) NGUYEN BA TRUNG ­ 2016 Thí dụ: expenditure.wf NGUYEN BA TRUNG ­ 2016 V Test the simple linear combination the parameters (1) Assume that we of want to H test: :β =β (2). Compute t­statistics as: v  Where: Se( βˆ1 − βˆ2 ) = H1 : β1 β2 βˆ1 − βˆ2 t= se( βˆ1 − βˆ2 ) ( se(βˆ ) ( − cov( βˆ1 , βˆ2 ) + se( βˆ2 (3). With α = 5%, search to find: tα/2(n­k­1) (4). Decision: If |t| > tα/2(n­k­1), We can reject H0 NGUYEN BA TRUNG ­ 2016 ) Example:Twoyears.wf • We have the following model: log(wage) = β + β1 jc + β 2univ + β exp er + u Where:  Jc: trình độ cao đẳng (2 năm đại học)  v  Unive: trình độ đại học (4 năm đại học)  v  Exper: kinh nghiệm (tháng) Question: whether wage differs between who those graduated from  colleague and university?  NGUYEN BA TRUNG ­ 2016 v  From the estimate result, the different wage:  θ1 = βˆ1 − βˆ2 = −0.0102 v  Is this difference statistically significant?  NGUYEN BA TRUNG ­ 2016 H : β1 = β (1) Set up the hypothesis: H1 : β1 < β v v  Suppose that: ˆ − βˆ ) = Se ( β  This implies:  = ( se( βˆ1 ) ( − cov( βˆ1 , βˆ2 ) + se( βˆ2 (0.0068) + (0.0023) = 0.00717 βˆ1 − βˆ2 0.0102  (2). Then we have: t = se( βˆ − βˆ ) = 0.00717 = −1.42 (3). With α = 5%, t0.025(6759) = 1.96 (4). Because |t| = 1.42                           , we can reject H0 NGUYEN BA TRUNG ­ 2016 r Example: BWGHT.wf • Compute F-statistics as: RSS r − RSS ur / q (0.0387 − 0.0364) / F= = = 1.42 RSS ur / n − k − (1 − 0.0387) / (1185) v  Search to find:  F0.05(2,1185) = 3 v  Because F =1.42  t /2(n-k-1) => Reject H0 If NGUYEN BA TRUNG ­ 2016 t t /2(n-k-1) => Accept... alternative hypothesis H1:  1 > 0 NGUYEN BA TRUNG ­ 2016 II. HYPOTHESES TESTING 2.2 Testing against One-Sided Alternatives: Left Hand (1) Side Suppose the hypothesis: H : β1 = H1 : β1 < (2).Compute t­statistics:... 2.5 Compute P-value for t-tests • • • • With confident level α, P-value is the probability of mistake if we reject the null hypothesis H0 P-value: Prob(|T|>|t|) Thus, if small P-value provides

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